Question

Cu electrode is dipped in 0.1 M of its solution at
25°C. Assuming salt to be 50% dissociated,
reduction potential of electrode is
[E°Cu/Cu2+ = -0.34 V]
(1) +0.374 V
(2) -0.374 V
(3) -0.302 V
(4) +0.302 V​

Answer 1

Answer:

the answer is +0.374 it is right